Wrong $f$! Let $g(x)=e^x$. Write down the series for $g(x)$. Evaluate at $x$, $-x$, $ix$, $-ix$. Add up. The coefficients of $x^{4k+i}$ are $0$ for $i=1,2,3$ and $\frac{4}{(4k)!}$ for $i=0$.
–
André NicolasOct 26 '12 at 14:46

5 Answers
5

Let $e(x)=\mathrm e^x=\displaystyle\sum\limits_{n\geqslant0}\frac{x^n}{n!}$. Note that, for every integer $n$, $\displaystyle\sum\limits_\zeta\zeta^n$ is $4$ when $n=0\pmod{4}$ and is $0$ otherwise, where the sum is over the set of roots $\{\zeta\in\mathbb C\mid \zeta^4=1\}=\{1,-1,\mathrm i,-\mathrm i\}$. Hence,
$$
\sum\limits_\zeta e(\zeta x)=\sum\limits_{n\geqslant0}\frac{x^n}{n!}\cdot\sum\limits_\zeta\zeta^n=4\cdot\sum\limits_{n\geqslant0}\frac{x^{4n}}{(4n)!}.
$$
In other words,
$$
\sum\limits_{n\geqslant0}\frac{x^{4n}}{(4n)!}=\frac{e(x)+e(-x)+e(\mathrm ix)+e(-\mathrm ix)}4=\frac{\cosh(x)+\cos(x)}2.
$$

Because the discrete Fourier transform of $(x_n)_{0\leqslant n\leqslant3}$ is $(y_k)_{0\leqslant k\leqslant3}$ defined by $y_k=\sum\limits_{n=0}^3x_n\zeta^{-kn}$. The beginning of the post is based on the fact that $(4,0,0,0)$ is the Fourier transform of $(1,1,1,1)$.
–
DidOct 26 '12 at 14:36

The fourth derivative of the function is the function itself. Then apply the well-known theory for solution of a linear ODE with constant coefficients. Of course you get the same answer as the other methods. But this method (find a DE it satisfies) can be applied in many cases where you want to sum a power series.

I'm not familiar with Laplace transform. Would you please tell me how I could understand it intuitively?
–
FrenzY DT.Oct 26 '12 at 14:50

@FrenzYDT: Laplace transform is a very useful technique and it has a lot of applications such as solving differential equations. You should practice it to learn it.
–
Mhenni BenghorbalOct 26 '12 at 15:41